Physics 211: Lecture 21 Todays Agenda

1
Physics 211 Lecture 21Todays Agenda

• Recap
• Beam Strings
• What if a string breaks?
• Ladder
• Hinged Beams
• Stability
• Fridge on truck
• Spool

2
Statics Review

• In general, we can use the two equations
• to solve any statics problem.
• When choosing axes about which to calculate
  torque, we can be clever and make the problem
  easy!!

3
Lecture 21, Act 1Statics

• In which of the static cases shown below is the
  tension in the supporting wire bigger? In both
  cases M is the same, and the blue strut is
  massless.

(a) case 1 (b) case 2
(c) same
T1
T2
300
300
L
L/2
M
M
case 1
case 2
4
Lecture 21, Act 1 Solution

• Consider the torque about the hinge between the
  strut the wall

due to gravity
due to wire
T1
300
L
does not depend on length of massless beam!
M
5
Lecture 21, Act 1 Solution

• The tension is the same in both cases.

T 2Mg
T 2Mg
300
300
L
L/2
M
M
case 1
case 2
6
Review Beam and Strings

• We can solve for the tensions in the strings in
  the following problem

T1
T2
M
x cm
L/2
L/4

• But what if a string breaks...

Mg
7
Beam and Strings...

• If the left string breaks, what is the initial
  acceleration of the CM?
• The beam will rotate about the axis at A.
• Using F ma in the y direction
• Mg - T Ma
• Figure out I about A
• I ICM Md2.

T
M
x A
x CM
d
Mg
8
Beam and Strings...

• Figure out ? about A
• ?? Mgd
• Now use ?? I??and a ?d

T
M
x A
x CM
d
a
9
Comment Center of Mass Statics

• The center of mass is at the point where the
  system balances!
• Sum of all gravitational torques about an axis
  through the CM 0!

d1
d2
CM
10
Lecture 21, Act 2Statics

• A (static) mobile hangs as shown below. The
  rods are massless and have lengths as indicated.
  The mass of the ball at the bottom right is 1kg.
• What is the total mass of the mobile?

(a) 5 kg (b) 6 kg (c)
7 kg
1 m
2 m
1 kg
1 m
3 m
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Lecture 21, Act 2 Solution

• First figure out M1.

M2
1 m
2 m
1 kg
M1
1 m
3 m
12
Lecture 21, Act 2 Solution

• First figure out M1.

• So the bottom part has a total mass of 4 kg.

2 kg
1 m
2 m
1 kg
3 kg
1 m
3 m
13
Lecture 21, Act 2 Solution

• Now figure out M2.

2 kg
1 m
2 m
4 kg
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Lecture 21, Act 2 Solution

• So the total mass of the mobile is
• 1 kg 3 kg 2 kg 6 kg

M2
1 m
2 m
1 kg
3 kg
1 m
3 m
15
Hinged Beams

• Consider a structure made from two beams,
  attached to each other and the wall with hinges

??
L
16
Hinged Beams...

• What we want to find
• (Ax,Ay) and (Bx,By) these are forces acting on
  beams at A B.These keep the beams from moving
• Which way are they pointing?
• What we know
• Sum of forces in x and y directions is zero
• Torque around any axis (A, B and C) is also zero
• WHY??
• No rotation! Its static

m2g
??
C
m1g
L
17
Hinged Beams...

• First use FNET ma in x and y directions
• x Ax Bx 0 Ax -Bx
• y Ay By (m1m2)g

Ay
(a)
Ax
(b)
Thats two equations, but we have four
unknowns... What about the actual directions of
Ax and Bx Which one points to R and L?
By
m2g
??
Bx
m1g
L
18
Hinged Beams...

• Now use some torque relationships.
• First, consider the torque on the whole structure
  about an axis through the hinge at B.

Ax
m2g
??
C
B
m1g
L
(No torques from forces at C, they cancel)
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Hinged Beams...

• If we knew something about Ay or By we would be
  done!
• Do the simplest thing we can think of!
• Consider the torque on the bottom beam about an
  axis through C

By
(d)
C
m1g
L
20
Hinged Beams...

• So we have the following equations
• Ax -Bx
• Ay By (m1m2)g

Ay
(a)
Ax
(b)
m2
(c)
By
??
(d)
Bx
m1
So why do we have two torque equations, now?
2 bodies
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Hinged Beams...

• And we solve these to get
• What happens if f is small?
• Y-forces unchanged
• X-forces are changed, then.

Ay
Ax
m2
By
??
Bx
m1
22
Lecture 21, Act 3Statics

• In which of the static cases shown below is the
  tension in the supporting wire bigger? In both
  cases the red strut has the same mass and length.

(a) case 1 (b) case 2
(c) same
T1
T2
300
300
L/2
L
case 1
case 2
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Lecture 21, Act 3 Solution

• Consider the total torque about an axis through
  the hinge

• The torque due to the wire must also be the same
  in both cases.

T1
T2
300
300
m
L
L/2
case 1
case 2
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Lecture 21, Act 3 Solution

• In case 1 the torque due to the wire is t1 T1
  x sin(30) x L

• In case 2 the torque due to the wire is t2 T2
  x sin(30) x L/2

• In order for these to be the same T2 2T1

T1
T2
300
300
m
L
L/2
mg
mg
case 1
case 2
25
Stability of a static structure.
Triangles vs. squares

• The triangular hinged structure we just studied
  is clearly rigid and stable (it does not tend to
  change shape).
• A hinged structure with more sides will NOT in
  general be rigid

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Stability of a static structure...

• Rigid construction generally consists of
  triangular elements.
• Look at bridges, etc.

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Stability of a static structure...
Tacoma Narrows

• Rigid construction generally consists of
  triangular elements.
• Look at bridges etc.

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Stability of a static structure...

• We can fill it in to make construction easier
  and the structure even more rigid

end view of I beam.
29
More on Stability

• Consider a truck moving a refrigerator of mass M.
  The CM of the fridge is a height h above the bed
  of the truck and the width of the fridge is 2w.
  If the truck is on a horizontal road, what is the
  maximum acceleration aM that the truck can have
  without tipping the fridge? (Assume the fridge
  does not slip).

2w
CM
h
aM
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Fridge...

• Suppose the trucks acceleration a is such that
  the fridge is just starting to tip. In this case
  the weight of the fridge is supported by a normal
  force acting only at the back corner.
• There must also be a frictional force acting to
  keep the fridge accelerating

w
CM
a
h
F Ma
N Mg
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Fridge...

• Since the fridge is not rotating, the sum of all
  the torques about an axis through the CM must be
  zero.
• Mgw Mah

w
CM
a
h
F Ma
N Mg
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Fridge...
Fridge on truck

• Since the torque due to the normal force cant be
  any bigger, if we increased the acceleration, the
  net torque would be non-zero, and the fridge
  would flip.
• This must be the maximum allowable acceleration!

w
CM
aM
h
F MaM
N Mg
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Lecture 21, Act 4Stick on the Water
Stick on the water

• Suppose one were to hang a stick from a platform
  with a string. The bottom of the stick rests on
  a platform which freely floats on the water.
  Which of the following configurations most
  rationally represents the equilibrium condition
  of this setup?

b)
a)
c)
x
CM
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Lecture 21, Act 4Solution

• Since the floating platform floats freely, there
  will be no x component to the force on the
  string, so the string must be vertical.

b)
a)
c)
x
CM
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More stability Spool on a rough surface.

• A spool (yo-yo) with inner radius a and outer
  radius b is at rest on a rough horizontal table.
  A string is wound around the inner radius, and
  extends behind the spool making an angle ? with
  the horizontal axis. There is tension T in the
  string. What is ? such that the spool does not
  move?

T
?
a
b
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Spool on a rough surface...

• Consider all of the forces acting tension T and
  friction f.
• Using FNET 0 in the x direction

T
?
Solving
a
b
f
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Spool on a rough surface...

• There is another (slick) way to see this
• Consider the torque about the point of contact
  between the spool and the ground. We know the net
  torque about this (or any other) point is zero.
• Since both Mg and f act through this point, they
  do not contribute to the net toque.
• Therefore the torque due to T must also be zero.
• Therefore T must actalong a line that
  passesthrough this point!

T
?
a
b
Mg
f
38
Spool on a rough surface...
Giant yo-yo

• So we can use geometry to get the same result.

T
?
a
?
b
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Recap of todays lecture

• Recap
• Beam Strings
• What if a string breaks?
• Ladder
• Hinged Beams
• Stability
• Fridge on truck
• Spool